Problem: Simplify and expand the following expression: $ \dfrac{2}{3t + 24}+ \dfrac{1}{5t - 15}- \dfrac{4t}{t^2 + 5t - 24} $
Answer: First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor a $3$ out of denominator in the first term: $ \dfrac{2}{3t + 24} = \dfrac{2}{3(t + 8)}$ We can factor a $5$ out of denominator in the second term: $ \dfrac{1}{5t - 15} = \dfrac{1}{5(t - 3)}$ We can factor the quadratic in the third term: $ \dfrac{4t}{t^2 + 5t - 24} = \dfrac{4t}{(t + 8)(t - 3)}$ Now we have: $ \dfrac{2}{3(t + 8)}+ \dfrac{1}{5(t - 3)}- \dfrac{4t}{(t + 8)(t - 3)} $ The least common multiple of the denominators is: $ 15(t + 8)(t - 3)$ In order to get the first term over $15(t + 8)(t - 3)$ , multiply by $\dfrac{5(t - 3)}{5(t - 3)}$ $ \dfrac{2}{3(t + 8)} \times \dfrac{5(t - 3)}{5(t - 3)} = \dfrac{10(t - 3)}{15(t + 8)(t - 3)} $ In order to get the second term over $15(t + 8)(t - 3)$ , multiply by $\dfrac{3(t + 8)}{3(t + 8)}$ $ \dfrac{1}{5(t - 3)} \times \dfrac{3(t + 8)}{3(t + 8)} = \dfrac{3(t + 8)}{15(t + 8)(t - 3)} $ In order to get the third term over $15(t + 8)(t - 3)$ , multiply by $\dfrac{15}{15}$ $ \dfrac{4t}{(t + 8)(t - 3)} \times \dfrac{15}{15} = \dfrac{60t}{15(t + 8)(t - 3)} $ Now we have: $ \dfrac{10(t - 3)}{15(t + 8)(t - 3)} + \dfrac{3(t + 8)}{15(t + 8)(t - 3)} - \dfrac{60t}{15(t + 8)(t - 3)} $ $ = \dfrac{ 10(t - 3) + 3(t + 8) - 60t} {15(t + 8)(t - 3)} $ Expand: $ = \dfrac{10t - 30 + 3t + 24 - 60t}{15t^2 + 75t - 360} $ $ = \dfrac{-47t - 6}{15t^2 + 75t - 360}$